I would like to start the beginning of this assignment by saying thank you to my professors of my last two algebra classes at Ashford. Although, this last class has been very frustrating for me with as many technical issues that I have had with the class and I am even ending the class on that note. I want to complete the class and may not be real happy with my grade I know that I have learned a lot of algrebra in the last 10 weeks. With that said this weeks assignment was on Composite and Inverse functions. The inverse of a function has all the same points as the original function, except that the x's and y's have been reversed. This is what they were trying to explain with their sets of points. In function composition, you're plugging entire functions in for the x. Below is my assignment problem for the week. Compositon and InverseWe defne The Following Functons:± (x) = 2x + 5 g(x)= x2 -3 h(x)=7-x/3
MAT 222 Week 5 DiscussionFor Week 5’s Discussion, we are to solve problems according to the last letter of our last name. Iwill be solving problems #16 and #50 on pages 708-711.#16) h(x) = |-2x|x|-2x|-5|-2(-5)| = 10-4|-2(-4)| = 8-3|-2(-3)| = 6-2|-2(-2)| = 4-1|-2(-1)| = 20|-2(0)| = 01|-2(1)| = 22|-2(2)|= 43|-2(3)| = 64|-2(4)| = 85|-2(5)| = 10This is a graph of an absolute value function. Plotting these points in a graph, they would form a v-shape. The v-shape would fall in the first and second quadrant. Since any real number can beused for x in h(x) =|-2x| and because the graph could keep extending to the left and right when adding more points, the domainis (−∞,∞). If we were looking at the graph, we would see that the vertex of the graph is (0, 0). Because |-2x| is never negative, the graph does not go below the x-axis. This means that the rangeof the graph would be [0,∞). This equation is a function. We can test this using the vertical line test. If we were to draw a straight line through any point it would only touch the graph once, so it passes the vertical line test.